Evaluate the following integrals: ^ 2x (3x+1)dx - Vidyadip

Question

Evaluate the following integrals:
$\int\text{e}^{2\text{x}}\sin(3\text{x}+1)\text{dx}$

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Answer

We have,
$\text{I}=\int\text{e}^{2\text{x}}\sin(3\text{x}+1)\text{dx}$
Let the first function be sin(3x + 1) and the second function be e^2x.
First we find the integral of the second function, i. e., $\int\text{e}^{2\text{x}}\text{dx}$.
$\int\text{e}^{2\text{x}}\text{dx}=\frac{1}{2}\text{e}^{2\text{x}}$
Now, using integration by parts, we get
$\text{I}=\sin(3\text{x}+1)\int\text{e}^{2\text{x}}\text{dx}-\int\Big[\Big(\frac{\text{d}(\sin(3\text{x}+1))}{\text{dx}}\Big)\Big]\text{dx}$
$=\frac{1}{2}\sin(3\text{x}+1)\text{e}^{2\text{x}}\text{dx}-\frac{3}{2}\int\big[\cos(3\text{x}+1)\text{e}^{2\text{x}}\big]\text{dx}$
$=\frac{1}{2}\sin(3\text{x}+1)\text{e}^{2\text{x}}-\frac{3}{2}\Big\{\cos(3\text{x}+1)\int\text{e}^{2\text{x}}\text{dx}-\int\Big[\Big(\frac{\text{d}(\cos(3\text{x}+1))}{\text{dx}}\Big)\int\text{e}^{2\text{x}}\text{dx}\Big]\text{dx}\Big\}$
$=\frac{1}{2}\sin(3\text{x}+1)\text{e}^{2\text{x}}-\frac{3}{4}\cos(3\text{x}+1)\text{e}^{2\text{x}}-\frac{9}{4}\text{I}+\text{C}$
$\text{I}+\frac{9}{4}\text{I}=\frac{1}{2}\sin(3\text{x}+1)\text{e}^{2\text{x}}-\frac{3}{4}\cos(3\text{x}+1)\text{e}^{2\text{x}}+\text{C}$
$\frac{13}{4}\text{I}=\frac{\text{e}^{2\text{x}}}{2}\big[\sin(3\text{x}+1)-\frac{3}{2}\cos(3\text{x}+1)\big]+\text{C}$
$\text{I}=\frac{2}{13}\text{e}^{2\text{x}}\big[\sin(3\text{x}+1)-\frac{3}{2}\cos(3\text{x}+1)\big]+\text{C}$
$=\frac{\text{e}^{2\text{x}}}{13}\big[2\sin(3\text{x}+1)-3\cos(3\text{x}+1)\big]+\text{C}$
Hence,
$\int\text{e}^{2\text{x}}\sin(3\text{x}+1)\text{dx}=\frac{\text{e}^{2\text{x}}}{13}\big[2\sin(3\text{x}+1)-3\cos(3\text{x}+1)\big]+\text{C}$

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